@AHSAN

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**Responses to your reply titled **__re-check__

**point [2]**

You wrote:

Now, use the solution of u(y) into Eq. 2 and solve it for T(y) against y with the help of Bcs given in Eq. 3.

Finally, use the solution of u(y) into Eq. 1 and solve for diff(p(x), x).

Equation **de1** writes symbolically** <A, u(y)> = p'(x)** where A is some differential operator and <A, u(y)> denotes its action on **u(y)**. Then **p'(x) **is obviously equal to **<A, u(y)> **which is a function of **y** alone;, let's say **F(y)**.

Thus **p(x) = w*F(y) + constant**.

**point [3]**

I wrote

Your bc are given at 3 points (-1, 0, 1) which is not supported by Maple.

Either you formulate the problem differently by considering a 1st problem in [-1..0], a second in [0..+1] and interface continuity equations... or you remove on point, for instance 0:

You replied

Ans: I checked the BCs and found correct.

Your answer "*I checked the BCs and found correct*" makes no sense.

Maybe you didn't understand my comment: **you cannot solve a 1D boundary value problem by setting boundary conditions at three different points**.

Either you remove one of these 3 points, for instance **y=0**... or you solve two coupled BVPs: one is a BVP in the domain** [-1, 0]** and the other in the domain **[0, +1]** and you have to write compatibility (contnuity) conditions at their interface **y=0**.

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**Responses to your reply titled **__look like this__

Please do not focus on the plots beat read carefully the conditions I use in my last attached file.

(__Note: had you correctly answered my initial comments things could be likely clerer... but you didn't__)

I already told you that **Br** is never defined in you data so I took arbitrarily Br=1, which could easily explain the differences between the **T curves**.

Concerning the differences between the** u-curves** : check the data you give in your attached file to those used to produce these last results.

As I simply use dsolve there is no reason ti think that the **u(y)** curves I provide I wrong

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**Responses to your reply titled ** **Solution verifiaction**

**[1]**

When you derive equation **5**, whose **rhs = diff(p(y), y)**, wrt **y**, you det a rhs equal to **0**: doesn't this mean that **diff(p(y), y) = 0 **and then that **p(y)=a+b*y** ?

**[2]**

Note that you write **lhs(eq 5) = diff(p(x), x) **which I thought fristly (and maybe wrongly) was a typo. The reason us that the **lhs** of eq **5** is a function of **y** alone while its **rhs** is a function of **x** alone.

To get **de4** you differentiate **de3** wrt **y** , which gives** rhs(de4) = 0 **.

**dsolve***ing* **de4** then givesautomatically a constant value for **diff(p(x), x)**.

~~Look~~ **CODE REMOVED BY THE AUTHOR see here** f~~or an explanation on a toy example.~~

**So it is completely impossible to retrieve your second plot: it is not something that can be deduced from your equations nor from functions psi(y) and T(y)**

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~~Here is a revised version of my work ~~ **CODE REMOVED BY THE AUTHOR see here**

You are probably going to say that the results this file contains are not the same than those you presented before. Nevertheless I advice you to compare the equations you wrote to those where "your" results come from before posting a reply.

For now on you presented us a problem which is not correctly (well?) posed.